1. Ch. 8 Rotational Motion

2. 8-1 Torque A torque is an action that causes objects to rotate.
Torque is not the same thing as force.

3. Torque is created when the line of action of a force (an imaginary line that follows the direction of a force) does not pass through the center of rotation.

4. The lever arm is the distance between the point at which the force is applied and the center of rotation.

5. Lever arm length (m)
t = F x d
Torque (N.m)
Force (N)

6. When the force and lever arm are NOT perpendicular
t = F d sin θ

7. Example
A 20-centimeter wrench is used to loosen a bolt. The force is applied 0.20 m from the bolt.
It takes 50 N to loosen the bolt when the force is applied perpendicular to the wrench.
How much force would it take if the force was applied at a 30-degree angle from perpendicular?
1) You are asked to find the force.
2) You are given the force and lever arm for one condition.
3) The formula that applies is τ = rF.
4) Solve: The torque required to loosen the bolt τ = (50 N)(0.2 m) = 10 N.m
To get the same torque with a force applied at 30 degrees:
10 N.m = F × (0.2 m)cos30o = F
F = 10 N.m ÷ = 58 N. It takes a larger force.

8. Given:
d = 0.20 m
F1= 50 N
Θ1 = 90
Θ2 = 120

9. 8-2 Rotation & Inertia
Motion in which an entire object moves is called translation.
Motion in which an object spins is called rotation.
The point or line about which an object turns is its center of rotation.
An object can rotate and translate.

10. There are three different axes about which an object will naturally spin.
The point at which the three axes intersect is called the center of mass.

11. If an object is irregularly shaped, the center of mass can be found by spinning the object and finding the intersection of the three spin axes.
There is not always material at an object’s center of mass.

12. The center of gravity of an irregularly shaped object can be found by suspending it from two or more points.

13. For an object to remain upright, its center of gravity must be above its area of support.
The area of support includes the entire region surrounded by the actual supports.
An object will topple over if its center of mass is not above its area of support.

15. Rotational Dynamics
When an object is in rotational equilibrium, the net torque applied to it is zero.
Rotational equilibrium is often used to determine unknown forces.

16. Example
What are the forces (FA, FB) holding the bridge up at either end?

18. Example A boy and his cat sit on a seesaw.
The cat has a mass of 4 kg and sits 2 m from the center of rotation.
If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?
1) You are asked to find the boy’s lever arm.
2) You are given the two masses and the cat’s lever arm.
3) Torque, τ = rF weight: F = mg
In equilibrium the net torque must be zero.
4) Solve: For the cat,
τ = (2 m)(4 kg)(9.8 N/kg) = N-m
For the boy:
τ = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque must be zero.
d = 0
d = 0.16 m
The boy must sit quite close (16 cm) to the center.

19. Given:
m1 = 4 kg d1 = 2m m2 = 50 kg d2 = ?